To answer this question we will assume the following:
• The amount of gas is constant, that is, the moles do not change.
,• The pressure remains constant.
,• The gas behaves like an ideal gas.
With these assumptions, we can apply Charles's law which tells us:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]Where,
V1 is the initial volume of the gas in liters, 120mL=0.120L
T1 is the initial temperature in Kelvin, 27°C+273.15K=300.15K
V2 is the final volume of the gas in liters, 80.0mL=0.080L
T2 is the final temperature in Kelvin, unknown.
We clear T2 and replace the known data:
[tex]T_2=\frac{T_1}{V_1}\times V_2[/tex][tex]\begin{gathered} T_2=\frac{300.15K}{0.120L}\times0.080L=200.1K \\ T_2=-73.1\degree C \end{gathered}[/tex]Answer: The gas would have a volume of 80.0mL at -73.1°C
